Question

An urn contains 25 balls of which 10 balls bear a mark ‘X’ and the remaining 15 bear a mark ‘Y’. A ball is drawn at random from the urn, its mark is noted down and it is replaced. If 6 balls are drawn in this way, find the probability that
(i) all will bear ‘X’ mark.
(ii) not more than 2 will bear ‘Y’ mark.
(iii) at least one ball will bear ‘Y’ mark
(iv) the number of balls with ‘X’ mark and ‘Y’ mark will be equal.

Answer 1

Total number of balls in the urn = 25

Balls bearing mark ‘X’ = 10

 Balls bearing mark ‘Y’ = 15

p = P (ball bearing mark ‘X’) = 10/25=2/5

q = P (ball bearing mark ‘Y’) =15/25=3/5

Six balls are drawn with replacement. Therefore, the number of trials are Bernoulli trials. Let Z be the random variable that represents the number of balls with ‘Y’ mark on them in the trials.

(iii) P (at least one ball bears ‘Y’ mark) = P (Z ≥ 1) = 1 − P (Z = 0)